Factor the following expression: $ x^2 + 8xy + 12y^2 $
When we factor a polynomial of this form, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + ay)(x + by)&=&xx &+& xby + ayx &+& ayby \\ \\ &=& x^2 &+& {(a+b)}xy &+& {ab}y^2 \\ &\hphantom{=}& \hphantom{x^2} &\hphantom{+}& \hphantom{{8}xy} &\hphantom{+}& \hphantom{{12}y^2} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + ay)(x + by)}&\hphantom{=}&\hphantom{xx} &\hphantom{+}& \hphantom{xby + ayx} &\hphantom{+}&\hphantom{ayby} \\ &\hphantom{=}& \hphantom{x^2} &\hphantom{+}&\hphantom{{(a+b)}xy}&\hphantom{+}&\hphantom{{ab}y^2} \\ &=& x^2 &+& {8}xy &+& {12}y^2 \end{eqnarray} $ The coefficient on the $xy$ term is $8$ and the coefficient on the $y^2$ term is $12$ , so to reverse the steps above, we need to find two numbers that add up to $8$ and multiply to $12$ You can start by trying to guess which factors of $12$ add up to $8$ . In other words, you need to find the values for $a$ and $b$ that meet the following conditions: $ {a} + {b} = {8}$ $ {a} \times {b} = {12}$ If you're stuck, try listing out every single factor of $12$ and its opposite as $a$ in these equations, and see if it gives a value for $b$ that validates both conditions. For example, since $6$ is a factor of $12$ , try substituting $6$ for $a$ as well as $-6$ The two numbers $6$ and $2$ satisfy both conditions: $ {6} + {2} = {8} $ $ {6} \times {2} = {12} $ So we can factor the polynomial as $(x + 6y)(x + 2y)$.